If the tiger has consumed 20 pounds of food in 4 days that's \( \frac{20}{4} \) = 5 pounds of food per day. The tiger needs to consume 55 - 20 = 35 more pounds of food to reach 55 pounds total. At 5 pounds of food per day that's \( \frac{35}{5} \) = 7 more days.

A factorial is the product of an integer and all the positive integers below it. To solve a fraction featuring factorials, expand the factorials and cancel out like numbers:

\( \frac{6!}{3!} \)
\( \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 1} \)
\( \frac{6 \times 5 \times 4}{1} \)
\( 6 \times 5 \times 4 \)
120

To multiply terms with radicals, multiply the coefficients and radicands separately:

7\( \sqrt{4} \) x 4\( \sqrt{3} \)
(7 x 4)\( \sqrt{4 \times 3} \)
28\( \sqrt{12} \)

Now we need to simplify the radical:

28\( \sqrt{12} \)
28\( \sqrt{3 \times 4} \)
28\( \sqrt{3 \times 2^2} \)
(28)(2)\( \sqrt{3} \)
56\( \sqrt{3} \)

To subtract these fractions, first find the lowest common multiple of their denominators. The first few multiples of 3 are [3, 6, 9, 12, 15, 18, 21, 24, 27, 30] and the first few multiples of 5 are [5, 10, 15, 20, 25, 30, 35, 40, 45, 50]. The first few multiples they share are [15, 30, 45, 60, 75] making 15 the smallest multiple 3 and 5 share.

Next, convert the fractions so each denominator equals the lowest common multiple:

\( \frac{9 x 5}{3 x 5} \) - \( \frac{8 x 3}{5 x 3} \)

\( \frac{45}{15} \) - \( \frac{24}{15} \)

Now, because the fractions share a common denominator, you can subtract them:

\( \frac{45 - 24}{15} \) = \( \frac{21}{15} \) = 1\(\frac{2}{5}\)

First, solve for \( \left|y - 6\right| \):

\( \left|y - 6\right| \) + 7 = -3
\( \left|y - 6\right| \) = -3 - 7
\( \left|y - 6\right| \) = -10

The value inside the absolute value brackets can be either positive or negative so (y - 6) must equal - 10 or --10 for \( \left|y - 6\right| \) to equal -10:

So, y = 16 or y = -4.